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Protons are projected with an initial speed v_(i)=6kms^(-1) from a fileds-free region vec(E ) =-900hat jNC^(-1) present above the plane as shown in fig. The initial velocity vector of the protons makes an angle u with the plane. The proton are to hit a target that lies at a horizontal cross the plane and enter the electric field. Find the angle theta at which the protons must pass through the plane to strike the target. |
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Answer» Solution :`v_(i)=6.0xx10^(3)MS^(-1)` `a_(y)=(eE)/(m)=((1.6xx10^(-19))(900))/((1.6xx10^(-27)))=9.0xx10^(10)ms^(-2)` `R=(v_(i)^(2)sin 2 theta)/(a_(y))=2XX10^(-3)` `sin 2 theta =(1)/(2)` or `2 theta =30^(@)` or `theta =15^(@), 90^(@)-theta=75^(@)`. |
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